= Solution
Using <Cartesian coordinates>, the <Levi-Civita symbol> and the <Einstein summation convention>, the constant-vector assumption gives
$$
\begin{aligned}
[\nabla\times(\mathbf a\times\mathbf F)]_i
&=\epsilon_{ijk}\epsilon_{klm}a_l\partial_jF_m\\
&=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})a_l\partial_jF_m\\
&=a_i\partial_jF_j-a_j\partial_jF_i.
\end{aligned}
$$
Recognizing the <divergence> and <directional derivative> proves
$$
\boxed{\nabla\times(\mathbf a\times\mathbf F)=\mathbf a(\nabla\cdot\mathbf F)-(\mathbf a\cdot\nabla)\mathbf F.}
$$
Here the contractions use the <Kronecker delta>.
For an oriented piecewise smooth surface $S$ with boundary $C$, <Stokes theorem> states
$$
\oint_C\mathbf Q\cdot d\mathbf l=\int_S(\nabla\times\mathbf Q)\cdot d\mathbf S,
$$
where the orientations obey the <right-hand rule> and $\mathbf Q$ is continuously differentiable near $S$. Apply it to $\mathbf Q=\mathbf F\times\mathbf a$. The <scalar triple product> gives
$$
\begin{aligned}
\mathbf a\cdot\oint_Cd\mathbf l\times\mathbf F
&=\oint_C(\mathbf F\times\mathbf a)\cdot d\mathbf l\\
&=\int_S\left[(\mathbf a\cdot\nabla)\mathbf F-\mathbf a(\nabla\cdot\mathbf F)\right]\cdot d\mathbf S.
\end{aligned}
$$
The last integrand is $\mathbf a\cdot[(d\mathbf S\times\nabla)\times\mathbf F]$, as the component formula
$$
[(d\mathbf S\times\nabla)\times\mathbf F]_i=dS_j\partial_iF_j-dS_i\partial_jF_j
$$
shows. Since $\mathbf a$ is arbitrary, the <cross-product form of Stokes theorem> follows:
$$
\boxed{\oint_Cd\mathbf l\times\mathbf F=\int_S(d\mathbf S\times\nabla)\times\mathbf F.}
$$
The derivatives in the surface expression act on $\mathbf F$, not on the surface element.
For the specified square, the path is clockwise when viewed from positive $z$, so the compatible <unit normal> is $-\mathbf e_z$. On the first and fourth edges, $d\mathbf l$ and $\mathbf r$ are parallel, giving zero <cross product>. On the second edge, $y=1$ and $x$ runs from $0$ to $1$, so $d\mathbf l\times\mathbf r=\mathbf e_z\,dx$ contributes $\mathbf e_z$. On the third edge, $x=1$ and $y$ runs from $1$ to $0$, so the contribution is $-\mathbf e_z\int_1^0dy=\mathbf e_z$. Thus the <line integral> is $2\mathbf e_z$.
For an independent <surface integral>, $\partial_iF_j=\delta_{ij}$ and $\nabla\cdot\mathbf F=3$ for the <position vector> field. Its integrand is therefore $d\mathbf S-3d\mathbf S=-2d\mathbf S$. The square has unit area and downward orientation, giving
$$
\boxed{\oint_Cd\mathbf l\times\mathbf r=2\mathbf e_z=\int_S(-2d\mathbf S).}
$$
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